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I'm reading Hecht's Optics and I have a problem understanding a step in the derivation of the complex notation of waves

He writes that the wave equation for a harmonic wave can be written as $\Psi(x,t) = \operatorname{Re}(Ae^{i(\omega t -kx + \epsilon)})$ which is equal to $\Psi(x,t) = A \cos(\omega t -kx + \epsilon)$. And that step I understand.

However, next he writes that because of this the wave equation can be written as $\Psi(x,t) = Ae^{i(\omega t -kx + \epsilon)}$. And that step I don't understand. Why does he write that $\operatorname{Re}(Ae^{i(\omega t -kx + \epsilon)}) = Ae^{i(\omega t -kx + \epsilon)} $?

Edit: See my reason for believing that this is not a duplicate in the comments below

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  • $\begingroup$ I have seen that question when I googled for an answer, and although it may get me a little close to the answer, and don't feel like it answers my question. The trick seems to be to rewrite the Re(z) using eulers formula (first bullet on the second "yellow area" on the one you linked). But still I don't understand how to go from there to answer the question that I posted $\endgroup$ – user5744148 Jul 27 '19 at 11:16
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/316135/2451 , physics.stackexchange.com/q/53005/2451 , physics.stackexchange.com/q/77156/2451 and links therein. $\endgroup$ – Qmechanic Jul 27 '19 at 11:56
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    $\begingroup$ The term "wave equation" usually refers to the differential equation for wave propagation. What you have is an "equation of a wave", which is a solution of the wave equation. $\endgroup$ – user45664 Jul 27 '19 at 16:31
  • $\begingroup$ Yes, thanks @user45664 for that answer, I think it feels clearer now! $\endgroup$ – user5744148 Jul 28 '19 at 15:44
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Why does he write that $\operatorname{Re}(Ae^{i(\omega t -kx + \epsilon)}) = Ae^{i(\omega t -kx + \epsilon)} $?

He doesn't. As Hecht writes:

Henceforth, wherever it's convenient, we shall write the wave- function as $$\psi(x, t) = A e^{i(\omega t-kx+\epsilon)} = Ae^{i\varphi} \tag{2.37}$$ and utilize this complex form in the required computations. This is done to take advantage of the ease with which complex exponentials can be manipulated. Only after arriving at a final result, and then only if we want to represent the actual wave, must we take the real part. It has, accordingly, become quite common to write $\psi(x, t)$, as in Eq. (2.37), where it is understood that the actual wave is the real part.

At no point is there any pretense that $\operatorname{Re}(Ae^{i(\omega t -kx + \epsilon)})$ equals $Ae^{i(\omega t -kx + \epsilon)}$, because it doesn't.

Instead, one works with the complex representation $Ae^{i(\omega t -kx + \epsilon)}$, of which only the real part is the physical field, in the understanding that for linear wave propagation phemonema (i.e. everything that you might want to do, up to (but excluding) calculations of energies and intensities) the difference doesn't matter, and it is more convenient to work with the complex representation and then take the real part at the end as necessary.

... all of which has already been explained, in detail, in What is the physical significance of the imaginary part when plane waves are represented as $e^{i(kx-\omega t)}$?

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  • $\begingroup$ I'm sorry, but there is something I'm missing here. He essentially writes that $ Ae^{i(\omega t−kx+\epsilon)} = \Psi(x,t)= Re(Ae^{i(\omega t−kx+\epsilon)}) $ How does he not then claim what was in my original question, which you say he doesn't claim? $\endgroup$ – user5744148 Jul 27 '19 at 12:49
  • $\begingroup$ There is no such claim in the text, and he does not "essentially write that" it's true. The text is arguably less clear than what it could or should, but if you insist on holding on to your misreading despite all external arguments, then there's really no helping you. $\endgroup$ – Emilio Pisanty Jul 27 '19 at 13:02
  • $\begingroup$ @user5744148 there is no claim that $e^{i\phi}$ equals its real part, the text says that altough the wave is equal to $\mathrm{Re}(e^{i\phi})$, you can do all the computations using $e^{i\phi}$ instead and take the real part of the result, and you will get the same result as if you had done all the calculations using the real part from the beginning $\endgroup$ – user2723984 Jul 27 '19 at 13:08
  • $\begingroup$ Alright, I apparently had a bit of gap in the basic understanding, which is why I think I missunderstood you @EmilioPisanty. So just to clarify so I think I get it, the above are solutions to the differential wave equation. And we can just as well use the solution $ \Psi(x,t)=Ae^{i(\omegae t−kx+\epsilon)} because of the reason in the other posts mentioned above. Is that a correct interpretation? $\endgroup$ – user5744148 Jul 28 '19 at 15:46
  • $\begingroup$ Yes, mostly. We are interested in a certain solution of the wave equation (the real-valued cosine), but instead we work with a different solution which is easier to deal with and which trivially reduces to the solution we want whenever we choose to. Or, in other words, this question is a duplicate of the linked thread. $\endgroup$ – Emilio Pisanty Jul 28 '19 at 16:19

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